Evaluate the definite integral. $\int^{9}_{5}\left(\dfrac{x-4}{x^2}\right)\,dx = $ Choose 1 answer: Choose 1 answer: (Choice A) A $\ln\left(\dfrac95\right)-\dfrac{16}{45}$ (Choice B) B $\ln(5)$ (Choice C) C $\ln\left(4\right)+\dfrac{16}{45}$ (Choice D) D None of the above
Solution: First, simplify and use the power and natural log rules: $\begin{aligned}\int^{9}_{5}\left(\dfrac{x-4}{x^2}\right)\,dx ~&=~\int^{9}_{5}\left(\dfrac{x}{x^2}-\dfrac{4}{x^2}\right)\,dx \\&=~\int^{9}_{5}\left(\dfrac{1}{x}-4x^{-2}\right)\,dx \\&=\left(\ln(x)+4x^{-1}\right)\Bigg|^{9}_{5}\end{aligned}$ Second, plug in the limits of integration: $(\ln({9})+4\cdot{9}^{-1})-(\ln({5})+4\cdot{5}^{-1}) = \ln\left(\dfrac95\right)-\dfrac{16}{45}$. The answer: $\int^{9}_{5}\left(\dfrac{x-4}{x^2}\right)\,dx ~=~\ln\left(\dfrac95\right)-\dfrac{16}{45}$